2 5 Solving Equations Containing Integers Warm Up

2 -5 Solving Equations Containing Integers Warm Up Problem of the Day Lesson Presentation Course 2

2 -5 Solving Equations Containing Integers Warm Up Use mental math to find each solution. 1. 7 + y = 15 y=8 2. x ÷ 9 = 9 x = 81 3. 6 x = 24 x=4 4. x – 12 = 30 x = 42 Course 2

2 -5 Solving Equations Containing Integers Problem of the Day Zelda sold her wet suit to a friend for $156. She sold her tank, mask, and snorkel for $85 less than she sold her wet suit. She bought a used wet suit for $80 and a used tank, mask, and snorkel for $36. If she started with $0, how much money does she have left? $111 Course 2

2 -5 Solving Equations Containing Integers Learn to solve one-step equations with integers. Course 2

2 -5 Solving Equations Containing Integers To solve integer equations such as x – 2 = -3 you must isolate the variable on one side of the equation. One way to isolate the variable is to add opposites. Recall that the sum of a number and its opposite is 0. + – – – + + + – – -2 + 2 = 0 Course 2 –

2 -5 Solving Equations Containing Integers Helpful Hint 3 + (– 3) = 0 3 is the opposite of – 3. Course 2

2 -5 Solving Equations Containing Integers Additional Example 1 A: Solving Addition and Subtraction Equations Solve each question. Check each answer. – 6 + x = – 7 +6 +6 x = – 1 Check – 6 + x = – 7 ? – 6 + (– 1) = – 7 ? – 7 = – 7 Course 2 Add 6 to both sides to isolate the variable. Substitute – 1 for x in the original equation. True. – 1 is the solution to – 6 + x = – 7.

2 -5 Solving Equations Containing Integers Additional Example 1 B: Solving Addition and Subtraction Equations Solve each equation. Check each answer. p + 5 = – 3 + (– 5) +(– 5) p = – 8 Add – 5 to both sides to isolate the variable. Check p + 5 = – 3 ? – 8 + 5 = – 3 ? – 3 = – 3 Course 2 Substitute – 8 for p in the original equation. True. – 8 is the solution to p + 5 = – 3.

2 -5 Solving Equations Containing Integers Additional Example 1 C: Solving Addition and Subtraction Equations Solve each equation. Check each answer. y – 9 = – 40 +9 +9 Check y = – 31 y – 9 = – 40 ? – 31 – 9 = – 40 ? – 40 = – 40 Course 2 Add 9 to both sides to isolate the variable. Substitute – 31 for y in the original equation. True. – 31 is the solution to y – 9 = – 40.

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 1 A Solve each equation. Check each answer. – 3 + x = – 9 +3 +3 x = – 6 Check – 3 + x = – 9 ? – 3 + (– 6) = – 9 ? – 9 = – 9 Course 2 Add 3 to both sides to isolate the variable. Substitute – 6 for x in the original equation. True. – 6 is the solution to – 3 + x = – 9.

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 1 B Solve each equation. Check each answer. q + 2 = – 6 +(– 2) Check q = – 8 q + 2 = – 6 ? – 8 + 2 = – 6 ? – 6 = – 6 Course 2 Add – 2 to both sides to isolate the variable. Substitute – 8 for q in the original equation. True. – 8 is the solution to q + 2 = – 6.

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 1 C Solve each equation. Check each answer. y – 7 = – 34 +7 Check +7 y = – 27 y – 7 = – 34 ? – 27 – 7 = – 34 ? – 34 = – 34 Course 2 Add 7 to both sides to isolate the variable. Substitute – 27 for y in the original equation. True. – 27 is the solution to y – 7 = – 34.

2 -5 Solving Equations Containing Integers Additional Example 2 A: Solving Multiplication and Division Equations Solve each equation. Check each answer. b =6 – 5 b (– 5) – 5 = (– 5)6 b = – 30 Course 2 Multiply both sides by – 5 to isolate the variable.

2 -5 Solving Equations Containing Integers Additional Example 2 A Continued Check b =6 – 5 ? – 30 = 6 – 5 6 = 6 Course 2 Substitute – 30 for b in the original equation. True. – 30 is the solution to b = 6. – 5

2 -5 Solving Equations Containing Integers Additional Example 2 B: Solving Multiplication and Division Equations Solve each equation. Check each answer. – 400 = 8 y 8 8 – 50 = y Course 2 Divide both sides by 8 to isolate the variable.

2 -5 Solving Equations Containing Integers Additional Example 2 B: Solving Multiplication and Division Equations Check – 400 = 8 y Substitute – 50 for y in the original equation. ? – 400 = 8(– 50) – 400 = – 400 Course 2 True. – 50 is the solution to – 400 = 8 y.

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 2 A Solve each equation. Check each answer. c = – 24 4 4 c 4 = 4(– 24) Multiply both sides by 4 to isolate the variable. c = – 96 Course 2

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 2 A Continued Check c = – 24 4 Course 2 ? – 96 = – 24 4 Substitute – 96 for c in the original equation. – 24 = – 24 True. – 96 is the solution to c = – 24. 4

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 2 B Solve each equation. Check each answer. – 200 = 4 x 4 4 – 50 = x Course 2 Divide both sides by 4 to isolate the variable.

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Check It Out: Example 2 B Continued Check. – 200 = 4 x Substitute – 50 for x in the original equation. ? – 200 = 4(– 50) – 200 = – 200 Course 2 True. – 50 is the solution to – 200 = 4 x.

2 -5 Solving Equations Containing Integers Additional Example 3: Business Application In 2003, a manufacturer made a profit of $300 million. This amount was $100 million more than the profit in 2002. What was the profit in 2002? Let p represent the profit in 2002 (in millions of dollars). This year’s profit 300 is =100 million More than + 100 Last year’s profit p 300 = 100 + p – 100 200 = p The profit was $200 million in 2002. Course 2

2 -5 Solving Equations Containing Integers Check It Out: Example 3 This year the class bake sale made a profit of $243. This was an increase of $125 over last year. How much did they make last year? Let x represent the money they made last year. This year’s profit 243 is =100 million More than + 125 243 = 125 + x – 125 118 = x The class earned $118 last year. Course 2 Last year’s profit x

Equations Containing 2 -5 Solving Insert Lesson Title Here Integers Lesson Quiz Solve each equation. Check your answer. 1. – 8 y = – 800 2. x – 22 = – 18 3. – y 7 =7 100 4 – 49 4. w + 72 = – 21 – 93 5. Last year a phone company had a loss of $25 million. This year the loss is $14 million more last year. What is this years loss? $39 million Course 2